700=3t+4.9t^2

Solution for 700=3t+4.9t^2 equation:

700=3t+4.9t^2

We move all terms to the left:

700-(3t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-3t+700=0

a = -4.9; b = -3; c = +700;
Δ = b2-4ac
Δ = -32-4·(-4.9)·700
Δ = 13729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{13729}=\sqrt{1*13729}=\sqrt{1}*\sqrt{13729}=1\sqrt{13729}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1\sqrt{13729}}{2*-4.9}=\frac{3-1\sqrt{13729}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1\sqrt{13729}}{2*-4.9}=\frac{3+1\sqrt{13729}}{-9.8}
$